P23_001 - 1(a We note that the electric field points...

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Unformatted text preview: 1. (a) We note that the electric field points leftward at both points. Using F = q0 E , and orienting our x axis rightward (so ˆ points right in the figure), we find i F = +1.6 × 10−19 C −40 Nˆ i i = −6.4 × 10−18 N ˆ C which means the magnitude of the force on the proton is 6.4 × 10−18 N and its direction (−ˆ is i) leftward. (b) As the discussion in §23-2 makes clear, the field strength is proportional to the “crowdedness” of the field lines. It is seen that the lines are twice as crowded at A than at B , so we conclude that EA = 2EB . Thus, EB = 20 N/C. ...
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